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3 years ago

In a fuel cell there is direct conversion of mechanical energy into electricity.

Chemistry

1 answer:

rewona [7]3 years ago

5 0

Answer:

Explanation:

A fuel cell uses both hydrogen and oxygen as fuels. At the cathode oxygen is reduced to water according to the equation

O2+4H^+ +4e ------>2H2O

And hydrogen is oxidized at the anode:

2H2------>4H^+ +4e

Fuel cells produce only water as a by product hence they are environment friendly

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N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

3 years ago

Read 2 more answers

Convert 512 kilograms to milligrams. 0.000512 0.512 512,000 512,000,000

olga55 [171]

The answer is D! 512000000

5 0

3 years ago

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A chemist encounters an unknown metal. They drop the metal into a graduated cylinder containing water, and find the volume chang

Mashutka [201]

Answer:

The density of the metal is 0.561 g/mL

Explanation:

The computation of the density of the metal is shown below;

As we know that

The Density of the metal is

$= \frac{mass}{volume}$

where,

Mass = 4.9g

Change in volume = 6.9 mL

Now place these values to the above formula

So, the density of the metal is

$= \frac{4.9g}{6.9mL}$

= 0.561 g/mL

Hence, the density of the metal is 0.561 g/mL

We simply applied the above formula so that the correct density could arrive

5 0

3 years ago

Which of the following is the best choice for presenting the percent

Alik [6]

OD

hope this helps!!

3 0

3 years ago

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What volume is occupied by 0.104 mol of helium gas at a pressure of 0.91 atm and a temperature of 314 K ?

Aleksandr-060686 [28]

Answer:

The volume will be "2.95 L".

Explanation:

Given:

n = 0.104

p = 0.91 atm

T = 314 K

Now,

The Volume (V) will be:

= $\frac{nRT}{P}$

By putting the values, we get

= $\frac{0.104\times 0.0821\times 314}{0.91}$

= $\frac{2.6810}{0.91}$

= $2.95 \ L$

7 0

2 years ago